Question

An article a Florida newspaper reported on the topics that teenagers most want to discuss with their parents. The findings, the results of a poll, showed that 46% would like more discussion about the family's financial situation, 37% would like to talk about school, and 30% would like to talk about religion. These and other percentages were based on a national sampling of 531 teenagers. Estimate the proportion of all teenagers who want more family discussions about school. Use a 99% confidence level.

Answer 1

Answer:

Confidence Interval for the proportion of all teenagers who want more family discussions about school is ( 31.6% , 42.4% )

Step-by-step explanation:

Given:

Number of teenagers in the sample , n = 531

Percentage of people like more discussion about family's financial situation = 46%

Percentage of people like more discussion about school, $\hat{p}$ = 37%

Percentage of people like more discussion about religion = 30%

To find:  Confidence Interval for the proportion of all teenagers who want more family discussions about school

Level of confidence, α = 100 - 99 = 1% = 0.01

We know that

Critical Value for given level of significance, $z_c$ = 2.58

The Standard Deviation,

$\sigma_{\hat{p}}=\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=\sqrt{\frac{0.37\times(1-0.37)}{531}}=0.021$

The Required Confidence Interval is,

$\hat{p}\pm z_c.\sigma_{\hat{p}}=0.37\pm(2.58\times0.021)=0.37\pm0.0542$

$=(0.37-0.0542,0.37+0.542)=(0.316,0.424)$

Hence the Require confidence interval is ( 0.316 , 0.424 ).

Therefore, Confidence Interval for the proportion of all teenagers who want more family discussions about school is ( 31.6% , 42.4% )

Answer 2

From the information given, the 99% confidence interval estimate of the proportion of all teenagers who want more family discussions about school is (0.316, 0.424).

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $\frac{1+\alpha}{2}$.

The information given are as follows:

• Sample size of $n = 531$.
• Sample proportion of $\pi = 0.37$.

99% confidence level

So $\alpha = 0.99$, z is the value of Z that has a p-value of $\frac{1+0.99}{2} = 0.995$, so $z = 2.575$.

 

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.37 - 2.575\sqrt{\frac{0.37(0.63)}{531}} = 0.316$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.37 + 2.575\sqrt{\frac{0.37(0.63)}{531}} = 0.424$

The 99% confidence interval estimate of the proportion of all teenagers who want more family discussions about school is (0.316, 0.424).

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